The probability distribution of a discrete ra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Step $1$: Calculate $E(X) = \Sigma x P(X=x)$.$E(X) = (-1)(\frac{1}{3}) + (0)(\frac{1}{6}) + (1)(\frac{1}{6}) + (2)(\frac{1}{3}) = -\frac{1}{3} + 0

Vedclass · Accepted Answer

$\frac{113}{12}$

Vedclass · Answer

(A) Step $1$: Calculate $E(X) = \Sigma x P(X=x)$.$E(X) = (-1)(\frac{1}{3}) + (0)(\frac{1}{6}) + (1)(\frac{1}{6}) + (2)(\frac{1}{3}) = -\frac{1}{3} + 0 + \frac{1}{6} + \frac{2}{3} = \frac{-2+1+4}{6} = \frac{3}{6} = \frac{1}{2}$.Step $2$: Calculate $E(X^2) = \Sigma x^2 P(X=x)$.$E(X^2) = (-1)^2(\frac{1}{3}) + (0)^2(\frac{1}{6}) + (1)^2(\frac{1}{6}) + (2)^2(\frac{1}{3}) = \frac{1}{3} + 0 + \frac{1}{6} + \frac{4}{3} = \frac{2+1+8}{6} = \frac{11}{6}$.Step $3$: Calculate $\operatorname{var}(X) = E(X^2) - [E(X)]^2$.$\operatorname{var}(X) = \frac{11}{6} - (\frac{1}{2})^2 = \frac{11}{6} - \frac{1}{4} = \frac{22-3}{12} = \frac{19}{12}$.Step $4$: Calculate $6 \Sigma(x^2) P(X=x) - \operatorname{var}(X)$.$6 E(X^2) - \operatorname{var}(X) = 6(\frac{11}{6}) - \frac{19}{12} = 11 - \frac{19}{12} = \frac{132-19}{12} = \frac{113}{12}$.

$X=x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

The probability distribution of a discrete random variable $X$ is given below:
$X = x$ $-1$ $0$ $1$ $2$
$P(X = x)$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$

Then the value of $6 \Sigma(x^2) P(X=x) - \operatorname{var}(X) =$ ?

Similar Questions

$A$ coin is tossed three times. If $X$ denotes the absolute difference between the number of heads and the number of tails,then $P(X=1) = $

If $P(X=x)=k\left(\frac{3}{8}\right)^{X}, x=1,2,3, \ldots$ is the probability distribution function of a discrete random variable $X$,then $k=$

$A$ coin is tossed twice in succession. Let $X$ represent the number of tails in two tosses,then the probability distribution of $X$ is given by

If $P(X = x) = 5r^x$, $x = 1, 2, 3, \dots$ is the probability mass function of a discrete random variable $X$, then $r = $

If a random variable $X$ has the following probability distribution of $X$:

$X=x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$

Then $P(X \geqslant 6) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The probability distribution of a discrete random variable $X$ is given below: $X = x$$-1$$0$$1$$2$$P(X = x)$$\frac{1}{3}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{3}$Then the value of $6 \Sigma(x^2) P(X=x) - \operatorname{var}(X) =$ ?